a) Solve $2^x = 3$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2002 - Paper 1

To isolate $\cos x$, divide both sides by 2:

$\cos x = \frac{\sqrt{3}}{2}$ The general solution for this is:

$x = 2k\pi \pm \frac{\pi}{6}, \quad k \in \mathbb{Z}$

We set the polynomials equal to each other:

$x^3 - 2x^2 + a = (x+2)Q(x) + 3$ To find $a$, we can equate the coefficients. By substituting $x = -2$:

$(-2)^3 - 2(-2)^2 + a = 0 + 3$ This simplifies to:

$-8 - 8 + a = 3$ Thus,

$a = 19$

Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, we can rewrite the integral:

$\int_0^{\frac{\pi}{2}} \sin^2 4x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 - \cos 8x}{2} \, dx$ Calculating this gives:

$\frac{1}{2} \left[ x - \frac{\sin 8x}{8} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ Thus,

$2 \int_0^{\frac{\pi}{2}} \sin^2 4x \, dx = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$

Since $AB$ is a tangent to the circle at point T and $AC$ is a radius, by the Alternate Segment Theorem, $\angle ACB$ and $\angle TAB$ subtend the same arc AT. Thus, we have:

$\angle ACB = \beta$

Since $\angle ACB = \beta$ and $\angle AXY$ subtend the same arc AX, we have:

$\angle AXY = \beta$ Thus, we see that:

$\angle AXY = \angle ACB$ This implies that triangle AXY is isosceles, as two angles are equal.